Anyone know of a fast way to query multiple paths in Neo4j ?
Lets say I have movie nodes that can have a type that I want to match (this is psuedo-code)
MATCH
(m:Movie)<-[:TYPE]-(g:Genre { name:'action' })
OR
(m:Movie)<-[:TYPE]-(x:Genre)<-[:G_TYPE*1..3]-(g:Genre { name:'action' })
(m)-[:SUBGENRE]->(sg:SubGenre {name: 'comedy'})
OR
(m)-[:SUBGENRE]->(x)<-[:SUB_TYPE*1..3]-(sg:SubGenre {name: 'comedy'})
The problem is, the first "m:Movie" nodes to be matched must match one of the paths specified, and the second SubGenre is depenedent on the first match.
I can make a query that works using MATCH and WHERE, but its really slow (30 seconds with a small 20MB dataset).
The problem is, I don't know how to OR match in Neo4j with other OR matches hanging off of the first results.
If I use WHERE, then I have to declare all the nodes used in any of the statements, in the initial MATCH which makes the query slow (since you cannot introduce new nodes in a WHERE)
Anyone know an elegant way to solve this ?? Thanks !
You can try a variable length path with a minimal length of 0:
MATCH
(m:Movie)<-[:TYPE|:SUBGENRE*0..4]-(g)
WHERE g:Genre and g.name = 'action' OR g:SubGenre and g.name='comedy'
For the query to use an index to find your genre / subgenre I recommend a UNION query though.
MATCH
(m:Movie)<-[:TYPE*0..4]-(g:Genre { name:'action' })
RETURN distinct m
UNION
(m:Movie)-[:SUBGENRE]->(x)<-[:SUB_TYPE*1..3]-(sg:SubGenre {name: 'comedy'})
RETURN distinct m
Perhaps the OPTIONAL MATCH clause might help here. OPTIONAL MATCH beavior is similar to the MATCH statement, except that instead of an all-or-none pattern matching approach, any elements of the pattern that do not match the pattern specific in the statement are bound to null.
For example, to match on a movie, its genre and a possible sub-genre:
OPTIONAL MATCH (m:Movie)-[:IS_GENRE]->(g:Genre)<-[:IS_SUBGENRE]-(sub:Genre)
WHERE m.title = "The Matrix"
RETURN m, g, sub
This will return the movie node, the genre node and if it exists, the sub-genre. If there is no sub-genre then it will return null for sub. You can use variable length paths as you have above as well with OPTIONAL MATCH.
[EDITED]
The following MATCH clause should be equivalent to your pseudocode. There is also a USING INDEX clause that assumes you have first created an index on :SubGenre(name), for efficiency. (You could use an index on :Genre(name) instead, if Genre nodes are more numerous than SubGenre nodes.)
MATCH
(m:Movie)<-[:TYPE*0..4]-(g:Genre { name:'action' }),
(m)-[:SUBGENRE]->()<-[:SUB_TYPE*0..3]-(sg:SubGenre { name: 'comedy' })
USING INDEX sg:SubGenre(name)
Here is a console that shows the results for some sample data.
Related
I have a database that contains these four nodes:
Store, Guitar, GuitarModel, Accessory
*Guitar refers to a specific guitar that a person can own/play
optional match (a:Store), (b:Guitar), (c:GuitarModel), (d:Accessory)
where a.StoreNumber ="1234" and (a)-[:ContainsGuitar]->(b) and
(b)-[:IS_OF_MODEL]->(c) and
((d)-[:COMES_STANDARD]-(c) OR (d)-[:COMES_OPTIONAL]-(c) OR (d)-:COMES_OPTION_UPGRADE]-(c) OR (d)-[:COMES_UPGRADE]-(c))
return b.name, collect(d.name)
My issue right now is this query is pretty slow it takes about 120,000ms to perform.
I have 67,000 nodes and 131,000 relationships.
So am I doing something wrong that making this slow?
Do you have an index/constraint on :Store(StoreNumber) ?
Why are you only using an optional match ? You can combine MATCH & OPTIONAL MATCH
Why are you doing your pattern in the WHERE clause ? You should put it directly in a MATCH.
I think that your query creates a cartesian product between nodes, that's why it's so slow.
Can you try this query :
MATCH
(:Store { StoreNumber:"1234" })-[:ContainsGuitar]->(b)
RETURN
b.name,
[(b)-[:IS_OF_MODEL]->(:GuitarModel)-[:COMES_STANDARD|COMES_OPTIONAL|COMES_OPTION_UPGRADE|COMES_UPGRADE]-(d:Accessory) | d.name]
I want to create a map projection with node properties and some additional information.
Also I want to collect some ids in a collection and use this later in the query to filter out nodes (where ID(n) in ids...).
The map projection is created in an apoc call which includes several union matches.
call apoc.cypher.run('MATCH (n)-[:IS_A]->({name: "User"}) MATCH (add)-[:IS_A]->({name: "AdditionalInformationForUser"}) RETURN n{.*, info: collect(add.name), id: ID(n)} as nodeWithInfo UNION MATCH (n)-[:IS_A]->({Department}) MATCH (add)-[:IS_A]->({"AdditionalInformationForDepartment"}) RETURN n{.*, info: collect(add.name), id: ID(n)} as nodeWithInfo', NULL) YIELD value
WITH (value.nodeWithInfo) AS nodeWithInfo
WITH collect(nodeWithInfo.id) as nodesWithAdditionalInfosIds, nodeWithInfo
MATCH (n)-[:has]->({"Vacation"})
MATCH (u)-[:is]->({"Out of Order"})
WHERE ID(n) in nodesWithAdditionalInfosIds and ID(u) in nodesWithAdditionalInfosIds
return n, u, nodeWithInfo
This does not return anything because, when the where part is evaluated it doesn´t check "nodesWithAdditionalInfosIds" as a flat list but instead only gets one id per row.
The problem only exists because I am passing the ids (nodesWithAdditionalInfosIds) AND the nodeProjection (nodeWithInfo) on in the WITH clause.
If I instead only use the id collection and don´t use the nodeWithInfo projection the following adjustement works and returns my only the nodes which ids are in the id collection:
...
WITH collect(nodeWithInfo.id) as nodesWithAdditionalInfosIds
MATCH (n)-[:has]->({"Urlaub"})
MATCH (u)-[:is]->({"Out of Order"})
WHERE ID(n) in nodesWithAdditionalInfosIds and ID(u) in nodesWithAdditionalInfosIds
return n, u
If I just return the collection "nodesWithAdditionalInfosIds" directly after the WITH clause in both examples this gets obvious. Since the first one generates a flat list in one result row and the second one gives me one id per row.
I have the feeling that I am missing a crucial knowledge about neo4js With clause.
Is there a way I can pass on my listOfIds and use it as a flat list without the need to have an exclusive WITH clause for the collection?
edit:
Right now I am using the following workaround:
After I do the check on the ID of "n" and "u" I don´t return but instead keep the filtered "n" and "u" nodes and start a second apoc call that returns "nodeWithInfo" like before.
WITH n, u
call apoc.cypher.run('MATCH (n)-[:IS_A]->({name: "User"}) MATCH (add)-[:IS_A]->({name: "AdditionalInformationForUser"}) RETURN n{.*, info: collect(add.name), id: ID(n)} as nodeWithInfo UNION MATCH (n)-[:IS_A]->({Department}) MATCH (add)-[:IS_A]->({"AdditionalInformationForDepartment"}) RETURN n{.*, info: collect(add.name), id: ID(n)} as nodeWithInfo', NULL) YIELD value
WITH (value.nodeWithInfo) AS nodeWithInfo, n, u
WHERE nodeWithInfo.id = ID(n) OR nodeWithInfo.id = ID(u)
RETURN nodeWithInfo, n, u
This way I can return the nodes n, u and the additional information (to one of the nodes) per row. But I am sure there must be a better way.
I know ids in neo4j have to be used with care, if at all. In this case I only need them to be valid inside this query, so it doesn´t matter if the next time the same node has another id.
The problem is stripped down to the core problem (in my opinion), the original query is a little bigger with several UNION MATCH inside apoc and the actual match on nodes which ids are contained in my collection is checking for some more restrictions instead of asking for any node.
Aggregating functions, like COLLECT(), aggregate over a set of "grouping keys".
In the following clause:
WITH collect(nodeWithInfo.id) as nodesWithAdditionalInfosIds, nodeWithInfo
the grouping key is nodeWithInfo. Therefore, each nodesWithAdditionalInfosIds would always be a list containing one value.
And in the following clause:
WITH collect(nodeWithInfo.id) as nodesWithAdditionalInfosIds
there is no grouping key. Therefore, in this situation, nodesWithAdditionalInfosIds will contain all the nodeWithInfo.id values.
I have a graph database where there are user and interest nodes which are connected by IS_INTERESTED relationship. I want to find interests which are not selected by a user. I wrote this query and it is not working
OPTIONAL MATCH (u:User{userId : 1})-[r:IS_INTERESTED] -(i:Interest)
WHERE r is NULL
Return i.name as interest
According to answers to similar questions on SO (like this one), the above query is supposed to work.However,in this case it returns null. But when running the following query it works as expected:
MATCH (u:User{userId : 1}), (i:Interest)
WHERE NOT (u) -[:IS_INTERESTED] -(i)
return i.name as interest
The reason I don't want to run the above query is because Neo4j gives a warning:
This query builds a cartesian product between disconnected patterns.
If a part of a query contains multiple disconnected patterns, this
will build a cartesian product between all those parts. This may
produce a large amount of data and slow down query processing. While
occasionally intended, it may often be possible to reformulate the
query that avoids the use of this cross product, perhaps by adding a
relationship between the different parts or by using OPTIONAL MATCH
(identifier is: (i))
What am I doing wrong in the first query where I use OPTIONAL MATCH to find nonexistent relationships?
1) MATCH is looking for the pattern as a whole, and if can not find it in its entirety - does not return anything.
2) I think that this query will be effective:
// Take all user interests
MATCH (u:User{userId: 1})-[r:IS_INTERESTED]-(i:Interest)
WITH collect(i) as interests
// Check what interests are not included
MATCH (ni:Interest) WHERE NOT ni IN interests
RETURN ni.name
When your OPTIONAL MATCH query does not find a match, then both r AND i must be NULL. After all, since there is no relationship, there is no way get the nodes that it points to.
A WHERE directly after the OPTIONAL MATCH is pulled into the evaluation.
If you want to post-filter you have to use a WITH in between.
MATCH (u:User{userId : 1})
OPTIONAL MATCH (u)-[r:IS_INTERESTED] -(i:Interest)
WITH r,i
WHERE r is NULL
Return i.name as interest
In a Cypher query language for Neo4j, what is the difference between one MATCH clause immediately following another like this:
MATCH (d:Document{document_ID:2})
MATCH (d)--(s:Sentence)
RETURN d,s
Versus the comma-separated patterns in the same MATCH clause? E.g.:
MATCH (d:Document{document_ID:2}),(d)--(s:Sentence)
RETURN d,s
In this simple example the result is the same. But are there any "gotchas"?
There is a difference: comma separated matches are actually considered part of the same pattern. So for instance the guarantee that each relationship appears only once in resulting path is upheld here.
Separate MATCHes are separate operations whose paths don't form a single patterns and which don't have these guarantees.
I think it's better to explain providing an example when there's a difference.
Let's say we have the "Movie" database which is provided by official Neo4j tutorials.
And there're 10 :WROTE relationships in total between :Person and :Movie nodes
MATCH (:Person)-[r:WROTE]->(:Movie) RETURN count(r); // returns 10
1) Let's try the next query with two MATCH clauses:
MATCH (p:Person)-[:WROTE]->(m:Movie) MATCH (p2:Person)-[:WROTE]->(m2:Movie)
RETURN p.name, m.title, p2.name, m2.title;
Sure you will see 10*10 = 100 records in the result.
2) Let's try the query with one MATCH clause and two patterns:
MATCH (p:Person)-[:WROTE]->(m:Movie), (p2:Person)-[:WROTE]->(m2:Movie)
RETURN p.name, m.title, p2.name, m2.title;
Now you will see 90 records are returned.
That's because in this case records where p = p2 and m = m2 with the same relationship between them (:WROTE) are excluded.
For example, there IS a record in the first case (two MATCH clauses)
p.name m.title p2.name m2.title
"Aaron Sorkin" "A Few Good Men" "Aaron Sorkin" "A Few Good Men"
while there's NO such a record in the second case (one MATCH, two patterns)
There are no differences between these provided that the clauses are not linked to one another.
If you did this:
MATCH (a:Thing), (b:Thing) RETURN a, b;
That's the same as:
MATCH (a:Thing) MATCH (b:Thing) RETURN a, b;
Because (and only because) a and b are independent. If a and b were linked by a relationship, then the meaning of the query could change.
In a more generic way, "The same relationship cannot be returned more than once in the same result record." [see 1.5. Cypher Result Uniqueness in the Cypher manual]
Both MATCH-after-MATCH, and single MATCH with comma-separated pattern should logically return a Cartesian product. Except, for comma-separated pattern, we must exclude those records for which we already added the relationship(s).
In Andy's answer, this is why we excluded repetitions of the same movie in the second case: because the second expression from each single MATCH was using there the same :WROTE relationship as the first expression.
If a part of a query contains multiple disconnected patterns, this will build a cartesian product between all those parts. This may produce a large amount of data and slow down query processing. While occasionally intended, it may often be possible to reformulate the query that avoids the use of this cross product, perhaps by adding a relationship between the different parts or by using OPTIONAL MATCH (identifier is: (a)) .
IN short their is NO Difference in this both query but used it very carefully.
In a more generic way, "The same relationship cannot be returned more than once in the same result record." [see 1.5. Cypher Result Uniqueness in the Cypher manual]
How about this statement?
MATCH p1=(v:player)-[e1]->(n)
MATCH p2=(n:team)<-[e2]-(m)
WHERE e1=e2
RETURN e1,e2,p1,p2
I'd like to pull and combine data from several different paths that share a path at the beginning, not all of which might exist. For example, I'd like to do something like this:
MATCH (:Complex)-[:PATH]->(s:Somewhere)-[:FETCHING]->(data)
RETURN data.attribute
UNION ALL
MATCH (s)-[:OPTIONAL]->(o:OtherData)
RETURN o.attribute;
so that it doesn't retrace the path up to s. I can't actually do this, though, because UNION separates queries and the (s)-[:OPTIONAL] in the second part will match anything with an outgoing OPTIONAL relation; the s is a loose handle.
Is there a better way of doing this than repeating the path:
MATCH (:Complex)-[:PATH]->(s:Somewhere)-[:FETCHING]->(data)
RETURN data.attribute
UNION ALL
MATCH (:Complex)-[:PATH]->(s:Somewhere)-[:OPTIONAL]->(o:OtherData)
RETURN o.attribute;
I made a few attempts using WITH, but they all either caused the query to return nothing if any part failed, or I could not get them to line up into a single column and instead got rows with redundant data, or (with multiple, nested WITHs, which I'm not sure about the scoping of) just fetching everything.
Have you looked at the semantics of an optional match? So you can match to s, beyond s and your optional component. Something like:
MATCH (:Complex)-[:PATH]->(s:Somewhere)
MATCH (s)-[:FETCHING]->(data)
OPTIONAL MATCH (s)-[:OPTIONAL]->(otherData)
RETURN data.attribute, otherData.attribute
Sorry I missed the importance of a single column, is it really important?
You can gather the vaues into a single collection :
MATCH (:Complex)-[:PATH]->(s:Somewhere)
MATCH (s)-[:FETCHING]->(data)
OPTIONAL MATCH (s)-[:OPTIONAL]->(otherData)
RETURN [data.attribute] + COLLECT(otherData.attribute)
But doesn't this work for a single column:
MATCH (:Complex)-[:PATH]->(s:Somewhere)
MATCH (s)-[:FETCHING]->(data)
OPTIONAL MATCH (s)-[:OPTIONAL]->(otherData)
WITH [data.attribute] + COLLECT(otherData.attribute) as col
RETURN UNWIND col AS val